Difference between revisions of "2000 AMC 8 Problems/Problem 14"
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<math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math> | <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 9</math> | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/7an5wU9Q5hk?t=1552 | ||
==Solution== | ==Solution== | ||
Line 10: | Line 13: | ||
Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>. | Powers of <math>99</math> have the exact same property, so <math>99^{99}</math> also has a units digit of <math>9</math>. <math>9+9=18</math> which has a units digit of <math>8</math>, so the answer is <math>\boxed{D}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Using modular arithmetic: | ||
+ | <cmath>99 \equiv 9 \equiv -1 \pmod{10}</cmath> | ||
+ | |||
+ | Similarly, | ||
+ | <cmath>19 \equiv 9 \equiv -1 \pmod{10}</cmath> | ||
+ | |||
+ | We have | ||
+ | <cmath>(-1)^{19} + (-1)^{99} = -1 + -1 \equiv \boxed{(\textbf{D}) \ 8} \pmod{10}</cmath> | ||
+ | |||
+ | -ryjs | ||
+ | |||
+ | ==Solution 3== | ||
+ | Experimentation gives | ||
+ | <cmath>\text{any number ending with }9^{\text{something even}} = \text{has units digit }1</cmath> | ||
+ | |||
+ | <cmath>\text{any number ending with }9^{\text{something odd}} = \text{has units digit }9</cmath> | ||
+ | |||
+ | Using this we have | ||
+ | <cmath>19^{19} + 99^{99}</cmath> | ||
+ | <cmath>9^{19} + 9^{99}</cmath> | ||
+ | |||
+ | Both <math>19</math> and <math>99</math> are odd, so we are left with | ||
+ | <cmath>9+9=18,</cmath> which has units digit <math>\boxed{(\textbf{D}) \ 8}.</math> | ||
+ | -ryjs | ||
+ | |||
+ | ==Solution 4== | ||
+ | <math>19^{19}+99^{99}=</math>36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888 | ||
+ | 9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :-) If you are really stupid and have time (like about 3 hours), this is a real brainless bash. | ||
+ | |||
+ | --hefei417 | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2000|num-b=13|num-a=15}} | {{AMC8 box|year=2000|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Latest revision as of 05:23, 31 December 2022
Contents
Problem
What is the units digit of ?
Video Solution by OmegaLearn
https://youtu.be/7an5wU9Q5hk?t=1552
Solution
Finding a pattern for each half of the sum, even powers of have a units digit of
, and odd powers of
have a units digit of
. So,
has a units digit of
.
Powers of have the exact same property, so
also has a units digit of
.
which has a units digit of
, so the answer is
.
Solution 2
Using modular arithmetic:
Similarly,
We have
-ryjs
Solution 3
Experimentation gives
Using this we have
Both and
are odd, so we are left with
which has units digit
-ryjs
Solution 4
36972963764972677265718790562880544059566876428174110243025997242355257045527752342141065001012823272794097888
9548326540119429996769494359451621570193644014418071060667659303363419435659472789623878 which ends in 8. :-) If you are really stupid and have time (like about 3 hours), this is a real brainless bash.
--hefei417
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.